1.26 Upper and lower bounds
9 exam questions worth 25 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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A length is 6.3 cm to 1 decimal place. What is the upper bound?
Choose one answer.
Correct Not quite: the answer is 6.35 cm
Mark scheme
B: 6.35 cm
Model answer
6.35 cm (B). Half a unit is 0.05 and 6.3 + 0.05 = 6.35.
A mass is 450 g to the nearest 10 g. Write down the lower bound and the upper bound.
Mark scheme
B1: 445 g; B1: 455 g
Model answer
Lower bound 445 g and upper bound 455 g.
a = 3.2 and b = 4.5, both correct to 1 decimal place. (a) Work out the upper bound of a + b. (b) Work out the lower bound of b - a.
Mark scheme
M1: 3.25 + 4.55; A1: 7.8; M1: 4.45 - 3.25; A1: 1.2
Model answer
(a) 3.25 + 4.55 = 7.80. (b) 4.45 - 3.25 = 1.20.
A rectangle has length 5.3 cm and width 2.1 cm, both correct to 1 decimal place. Work out the upper bound of the area of the rectangle.
Mark scheme
B1: 5.35 and 2.15; M1: 5.35 × 2.15; A1: 11.5025
Model answer
UB of length 5.35, UB of width 2.15, so UB of area = 5.35 × 2.15 = 11.5025 cm².
For the same rectangle, work out the lower bound of the area. Hence write down the area to a suitable degree of accuracy.
Mark scheme
M1: 5.25 × 2.05; A1: 10.7625; B1: 10 cm² (1 s.f.) with both bounds rounding to 10
Model answer
LB = 5.25 × 2.05 = 10.7625. To 1 significant figure both 10.7625 and 11.5025 round to 10, so the area is 10 cm² (1 s.f.).
A runner travels 120 m, correct to the nearest 10 m, in 15.2 seconds, correct to 1 decimal place. Work out the upper bound of the speed. Give your answer to 3 significant figures.
Mark scheme
B1: UB distance 125 and LB time 15.15; M1: 125 ÷ 15.15; A1: 8.25 m/s
Model answer
Greatest speed = 125 ÷ 15.15 = 8.25 m/s (3 s.f.).
For the same runner, work out the lower bound of the speed. Give your answer to 3 significant figures.
Mark scheme
B1: LB distance 115 and UB time 15.25; M1: 115 ÷ 15.25; A1: 7.54 m/s
Model answer
Least speed = 115 ÷ 15.25 = 7.54 m/s (3 s.f.).
Explain why you divide the upper bound of the distance by the lower bound of the time to find the upper bound of the speed.
Mark scheme
B1: speed = distance ÷ time; B1: dividing by a smaller number gives a larger answer
Model answer
Speed = distance ÷ time. To make the speed as big as possible, use the largest distance and divide by the smallest time.
A metal block has a mass of 56 g, correct to the nearest gram, and a volume of 8.0 cm³, correct to 1 decimal place. Work out the density of the block to a suitable degree of accuracy. You must show your working. (density = mass ÷ volume)
Mark scheme
M1: UB = 56.5 ÷ 7.95; A1: 7.107...; M1: LB = 55.5 ÷ 8.05 = 6.894...; A1: both round to 7 to 1 s.f., so density = 7 g/cm³ (1 s.f.)
Model answer
UB density = 56.5 ÷ 7.95 = 7.107... LB density = 55.5 ÷ 8.05 = 6.894... To 1 s.f. both are 7, so the density is 7 g/cm³ to 1 significant figure.
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